2019 AMC 10B Problems/Problem 10
- The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.
Problem
In a given plane, points and are units apart. How many points are there in the plane such that the perimeter of is units and the area of is square units?
Solution 1
Notice that whatever point we pick for , will be the base of the triangle. Without loss of generality, let points and be and , since for any other combination of points, we can just rotate the plane to make them and under a new coordinate system. When we pick point , we have to make sure that its -coordinate is , because that's the only way the area of the triangle can be .
Now when the perimeter is minimized, by symmetry, we put in the middle, at . We can easily see that and will both be . The perimeter of this minimal triangle is , which is larger than . Since the minimum perimeter is greater than , there is no triangle that satisfies the condition, giving us .
~IronicNinja
Solution 2
Without loss of generality, let be a horizontal segment of length . Now realize that has to lie on one of the lines parallel to and vertically units away from it. But is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, . Dropping altitude , we have a right triangle with hypotenuse and leg , which is clearly impossible, again giving the answer as .
Solution 3
Area = , perimeter = , semiperimeter , , and .
Using the generic formula for triangle area using semiperimeter and sides , , and , area = . (Heron's formula)
.
Square both sides, divide by and expand the polynomial to get .
and the discriminant is so no real solutions.
Video Solution
~IceMatrix
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.